bzoj1089: [SCOI2003]严格n元树

来源：互联网发布：中国那个朝代最弱知乎编辑：程序博客网时间：2024/06/04 18:40

传送门
设f[i]表示深度小于等于i的树的个数。
得到公式
f[i]=f[i-1]^n+1
答案就是f[d]-f[d-1];
高精度水过。

type arr=array [0..5005] of int64;var  f:array [0..20] of arr;  n,d,i:longint;procedure cheng(a,b:arr; var c:arr);  var i,j:longint; x:int64;  begin    fillchar(c,sizeof(c),0);    for i:=1 to a[0] do begin      x:=0;      for j:=1 to b[0] do begin        x:=c[i+j-1]+x+a[i]*b[j];        c[i+j-1]:=x mod 100000000;        x:=x div 100000000;      end;      c[i+b[0]]:=x;    end;    if (c[a[0]+b[0]]<>0) then c[0]:=a[0]+b[0] else c[0]:=a[0]+b[0]-1;  end;procedure mi(x:longint);  begin    if (x=1) then exit;    mi(x div 2);    cheng(f[i],f[i],f[i]);    if x mod 2=1 then cheng(f[i],f[i-1],f[i]);    exit;  end;procedure jiayi;  var j:longint;  begin    inc(f[i,1]);    for j:=1 to f[i,0] do begin      inc(f[i,j+1],f[i,j] div 100000000); f[i,j]:=f[i,j] mod 100000000;    end;    if (f[i,f[i,0]+1]<>0) then inc(f[i,0]);  end;procedure jian;  var i:longint;  begin    for i:=1 to f[d,0] do f[d,i]:=f[d,i]-f[d-1,i];    for i:=1 to f[d,0] do      if f[d,i]<0 then begin inc(f[d,i],100000000); dec(f[d,i+1]); end;    while (f[d,0]<>1) and (f[d,f[d,0]]=0) do dec(f[d,0]);  end;procedure print;  var i:longint;  begin    write(f[d,f[d,0]]);    for i:=f[d,0]-1 downto 1 do begin      if f[d,i]<10000000 then write(0);      if f[d,i]<1000000 then write(0);      if f[d,i]<100000 then write(0);      if f[d,i]<10000 then write(0);      if f[d,i]<1000 then write(0);      if f[d,i]<100 then write(0);      if f[d,i]<10 then write(0);      write(f[d,i]);    end;  end;begin  read(n,d);  if d=0 then begin write(1); exit; end;  f[0,0]:=1; f[0,1]:=1;  for i:=1 to d do begin    f[i]:=f[i-1];    mi(n);    jiayi;  end;  jian;  print;end.

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